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1. SkyCiv RC设计
2. 板材设计模块
3. 澳大利亚标准 AS3600 楼板设计实例及与 SkyCiv 的比较

# 澳大利亚标准 AS3600 楼板设计实例及与 SkyCiv 的比较

## 标准考虑的平板系统    • 条款 6.10.2: 连续梁和单向板
• 条款 6.10.3: 四侧支撑的双向板
• 条款 6.10.4: 具有多个跨度的双向板 If you are new at SkyCiv, Sign up and test the software yourself!

## One-Way Slab Design Example

Shown below is the small building and the slabs we will design The plan dimensions are shown at next For the slab example, in summary, the material, elements properties, and loads to consider :

• Slab type classification: Oneway behaviour $$\压裂{L_2}{L_1} > 2 ; \压裂{14米}{6米}=2.33 > 2.00$$ 好的!
• Building occupation: Residential use
• Slab thickness $$t_{平板}=0.25m$$
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% $$\rho_w = 24 \压裂{千牛}{m^3} + 0.6 \压裂{千牛}{m^3} \次 0.5 = 24.3 \压裂{千牛}{m^3}$$
• Concrete characteristic compressive strength at 28 天 $$f'c = 25 兆帕$$
• Concrete Modulus of Elasticity by Australian Standard $$E_c = 26700 兆帕$$
• Slab Self-Weight $$Dead = \rho_w \times t_{平板} = 24.3 \压裂{千牛}{m^3} \times 0.25m = 6.075 \压裂 {千牛}{m^2}$$
• Super-imposed dead load $$SD = 3.0 \压裂 {千牛}{m^2}$$
• 活荷载 $$L = 2.0 \压裂 {千牛}{m^2}$$

### Hand calculation according to AS3600 Standard

• Dead load, $$g = (3.0 + 6.075) \压裂{千牛}{m^2} \次 1 m = 9.075 \压裂{千牛}{米}$$
• 活荷载, $$q = (2.0) \压裂{千牛}{m^2} \次 1 m = 2.0 \压裂{千牛}{米}$$
• Ultimate load, $$Fd = 1.2\times g + 1.5\times q = (1.2\次 9.075 + 1.5\次 2.0)\压裂{千牛}{米} =13.89 \frac{千牛}{米}$$

Using the simplified method specified by the standard, 第一, it is a must to comply with the following restrictions:

• $$\压裂{L_i}{L_j} \这 1.2 . \压裂{6米}{6米} =1 < 1.2$$. 好的!
• Load has to be uniform. 好的!
• $$q \le 2g. q=2 \frac{千牛}{米} < 18.15 \压裂{千牛}{米}$$. 好的!
• The slab cross-section has to be uniform. 好的!.

Recommended minimum thickness, d

$$d \ge \frac{L_{fe}}{{k_3}{k_4}{\sqrt{\压裂{\压裂{\三角洲}{L_{ef}}{E_c}}{F_{d, ef}}}}}$$

• $$k_3 = 1.0; k_4 = 1.75$$
• $$\压裂{\三角洲}{L_{ef}}=1/250$$
• $$E_c = 27600 兆帕$$
• $$F_{d,ef} = (1.0 +钢底板设计欧洲规范{cs})\times g + (\psi_s + 钢底板设计欧洲规范{cs}\times \psi_1) \times q=(1.0+0.8)\次 9.075 + (0.7+0.8\次 0.4)\次 2 = 18.375 kPa$$
• $$\psi_s = 0.7$$ Live-load short-term factor
• $$\psi_1 = 0.4$$ Live-load long-term factor
• $$钢底板设计欧洲规范{cs} = 0.8$$

$$d \ge \frac{5.50米}{{1.0}\次 {1.75}{\sqrt{\压裂{\压裂{1}{250}\次{27600 \times 10^3 kPa}}{18.375 千帕}}}} \ge 0.173m. d = 0.25m > 0.173米$$ 好的!

Once we demonstrate that constraints are satisfied, the bending moment is calculated using the expression: $$M=\alpha \times F_d \times L_n^2$$ 哪里 $$\alpha$$ is a constant defined in the following figure. • (一个) Case of slabs and beams on girder support
• (b) For continuous beam support only
• (C) Where Class L reinforcement is used
• $$L_n$$ is the unitary strip span
• $$F_d$$ is the gravitational factored load

For the slab example, we have to use case (一个) because the slab rests on stiff girders. It will be explained only one case and the rest will show in the following table. We include also the steel reinforcement area calculation.

• $$M={\α} {F_d}{L_n^2}={-\压裂{1}{24}}\次 {13.89 \压裂{千牛}{米}}\次 (6米-0.5米)可以假设为 – 17.51{千牛}{米}$$
• Cover = 20mm (A minimum of 10mm is needed for fire resistance period of 60 分钟).
• $$d = t_{平板} – 覆盖 – \压裂{BarDiameter}{2} = 250mm – 20毫米 – 6mm = 224mm$$
• $$\alpha_2 = 1.0-0.003 f’c = 1.0-0.003\times 25 = 0.925 (0.67 \le \alpha_2 \le 0.85)$$ 从而, we select $$\alpha_2 = 0.85$$
• $$\xi = \frac{\alpha_2\times f’c}{F_{他的}} = frac{0.85\次 25 兆帕}{500 兆帕} = 0.0425$$
• $$\rho_t = \xi – \sqrt{{\习}^ 2 – \压裂{{2}{\习}{中号}}{{\φ}{b}{d^2}{F_{他的}}}} = 0.0425 – \sqrt{{0.0425}^2-\frac{2\times 0.0425\times 17.51{千牛}{米}}{{0.8}\次 {1米}\次 {{(0.224米)^ 2}} \次 {500\次 {10^ 3}千帕}}}=0.0008814$$
• $$\gamma= 1.05-0.007 f’c = 1.05-0.007\times 25 = 0.875 (0.67 \le \gamma \le 0.85)$$ 从而, we select $$\gamma = 0.85$$
• $$k_u = \frac{\rho_t \times f_{他的}}{0.85\times \gamma \times f’c}= 分数{0.0008814\次 500 兆帕}{0.85\次 0.85 \次 25 兆帕} =0.0244$$
• $$\phi = 1.19 – \压裂{13\钢底板设计欧洲规范{u0}}{12} = 1.19 – \压裂{13\次 0.0244}{12} = 1.164 (0.6 \le \phi \le 0.8)$$ 从而, we select $$\phi = 0.8$$. 好的!.
• $$\o{Ť,分} = 0.20 {(\压裂{d}{d})^ 2}{(\压裂{F'_{ct,F}}{F_{他的}})} = 0.20 \次 (\压裂{0.25米}{0.224米})^2 \times \frac{0.6\次 sqrt{25兆帕}}{500 兆帕} = 0.0015$$
• $$一个_{圣}=max(\o{Ť,分}, \rho_t)\times b \times d = max(0.0015,0.0008814)\次 1000 mm \times 224 mm = 334.82 mm^2$$
$$\alpha$$ and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$\alpha$$ 值 -$$\压裂{1}{24}$$ $$\压裂{1}{11}$$ -$$\压裂{1}{10}$$ $$\压裂{1}{10}$$ $$\压裂{1}{16}$$ $$\压裂{1}{11}$$
M value -17.51 38.20 -42.02 42.02 26.26 38.20
$$\rho_t$$ 0.0008814 0.001948 0.002148 0.002148 0.00133 0.001948
ku 0.0244 0.0539 0.0594 0.0594 0.0368 0.05391
$$\phi$$ 0.8 0.8 0.8 0.8 0.8 0.8
$$一个_{圣} {mm^2}$$ 334.82 436.31 481.099 481.099 334.8214 436.3100

After the steel rebar area calculation, you can define the detailing (the actual way to place the reinforcement into the slab). As help for your knowing, we share the following image, which indicates the rebar location for positive and negative moments: If you are new at SkyCiv, Sign up and test the software yourself!

### SkyCiv S3D Plate Design Module Results

In the first view, we will show some images for the modeling and structural analysis of the example in S3D. We recommend you read about modeling in SkyCiv in the following links How to model plates?ACI Slab Design Example with SkyCiv. Before analyzing the model, we must define a plate mesh size. Some references (2) recommend a size for the shell element of 1/6 of the short span or 1/8 of the long span, the shorter of them. Following this value, 我们有 $$\压裂{L2}{6}= 分数{6米}{6} = 1m$$ 要么 $$\压裂{L1}{8}= 分数{14米}{8}=1.75m$$; we take 1m as a maximum recommended size and 0.50m applied mesh size. Once we improved our analytical structural model, we run a linear elastic analysis. When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of $$\压裂{大号}{250}= 分数{6000毫米}{250}=24.0 mm$$. Comparing the maximimum vertical displacement against the code referenced value, the slab’s stiffness is adequate. $$4.822 毫米 < 24.00mm$$.

The maximum moments in the slab’s spans are located for positive in the center and for negative at the exterior and interior supports. Let’s see these moments values in the following images.  Plate element local axes are indicated below. For more details about automated reinforced slab design, see our documentation Plates in SkyCiv.    ### Results comparison

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis (local axes “2”) and handcalculations.

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$一个_{圣, HandCalcs} {mm^2}$$ 334.82 436.31 481.099 481.099 334.8214 436.3100
$$一个_{圣, S3D} {mm^2}$$ 285.13 313.00 427.69 427.69 313.00 427.69
$$\三角洲_{dif}$$ (%) 14.84 28.262 11.101 11.101 6.517 1.986

We can see that the results of the values are very close to each other. This means the calculations are correct!

## Two-way Slab Design Example The plan dimensions are shown at next For the slab example, in summary, the material, elements properties, and loads to consider :

• Slab type classification: Twoway behaviour $$\压裂{L_2}{L_1} \这 2 ; \压裂{7米}{6米}=1.167 < 2.00$$ 好的!
• Building occupation: Residential use
• Slab thickness $$t_{平板}=0.25m$$
• Reinforced concrete density assuming a steel reinforcement ratio of 0.5% $$\rho_w = 24 \压裂{千牛}{m^3} + 0.6 \压裂{千牛}{m^3} \次 0.5 = 24.3 \压裂{千牛}{m^3}$$
• Concrete characteristic compressive strength at 28 天 $$f'c = 25 兆帕$$
• Concrete Modulus of Elasticity by Australian Standard $$E_c = 26700 兆帕$$
• Slab Self-Weight $$Dead = \rho_w \times t_{平板} = 24.3 \压裂{千牛}{m^3} \times 0.25m = 6.075 \压裂 {千牛}{m^2}$$
• Super-imposed dead load $$SD = 3.0 \压裂 {千牛}{m^2}$$
• 活荷载 $$L = 2.0 \压裂 {千牛}{m^2}$$

### Hand calculation according to AS3600 Standard

• Dead load, $$g = (3.0 + 6.075) \压裂{千牛}{m^2} \次 1 m = 9.075 \压裂{千牛}{米}$$
• 活荷载, $$q = (2.0) \压裂{千牛}{m^2} \次 1 m = 2.0 \压裂{千牛}{米}$$
• Ultimate load, $$Fd = 1.2\times g + 1.5\times q = (1.2\次 9.075 + 1.5\次 2.0)\压裂{千牛}{米} =13.89 \frac{千牛}{米}$$

Design moments and coefficients  Edge Condition Short-span coefficients ($$\beta_x$$) Long-span coefficients ($$\beta_y)$$ all values of $$\压裂{L_y}{L_x}$$

1.0 1.1 1.2 1.3 1.4 1.5 1.75 $$\葛 2.0$$
1. Four edges continuous 0.024 0.028 0.032 0.035 0.037 0.040 0.044 0.048 0.024
2. One short edge discontinuos 0.028 0.032 0.036 0.038 0.041 0.043 0.047 0.050 0.028
3. One long edge discontinous 0.028 0.035 0.041 0.046 0.050 0.054 0.061 0.066 0.028
4. Two short edges discontinous 0.034 0.038 0.040 0.043 0.045 0.047 0.050 0.053 0.034
5. Two long edges discontinous 0.034 0.046 0.056 0.065 0.072 0.078 0.091 0.100 0.034
6. Two adjacent edges discontinous 0.035 0.041 0.046 0.051 0.055 0.058 0.065 0.070 0.035
7. Three edges discontinuous (one long edge continuous) 0.043 0.049 0.053 0.057 0.061 0.064 0.069 0.074 0.043
8. Three edges discontinuous (one short edge continous) 0.043 0.054 0.064 0.072 0.078 0.084 0.096 0.105 0.043
9. Four edges discontinuos 0.056 0.066 0.074 0.081 0.087 0.093 0.103 0.111 0.056

The following image explain the all nine cases that the table above refers Design moments for central region (案件 6 Two adjacent edges discontinuous) :

• $$L_x = 6m, L_y=7m, \压裂{L_y}{L_x} = frac{7米}{6米}= 1.167$$ Values to be linearly interpolated
• Positives:
• $$M_x = {\beta_x}{F_d}{L_x^2} = {0.04435}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}=22.177 kNm$$
• $$M_y = {\beta_y}{F_d}{L_x^2} ={0.035}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}=17.501 kNm$$
• Negatives exterior span:
• $$M_{x1,A} = -\lambda_e \times M_x = -0.5 \次 22.177 kNm = – 11.089 kNm$$
• $$M_{y1,A} = -\lambda_e \times M_y = -0.5 \次 17.501 kNm = -8.751 千牛·米$$
• Negatives interior span:
• $$M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \次 22.177 kNm = – 29.495 kNm$$
• $$M_{y1, 乙} = -\lambda_{1和} \times M_y = -1.33 \次 17.501 kNm = -23.276 千牛·米$$

Design moments for central region (案件 3 One long edge discontinous) :

• $$L_x = 6m, L_y=7m, \压裂{L_y}{L_x} = frac{7米}{6米}= 1.167$$ Values to be linearly interpolated
• Positives:
• $$M_x = {\beta_x}{F_d}{L_x^2} = {0.03902}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}= 19.512 kNm$$
• $$M_y = {\beta_y}{F_d}{L_x^2} ={0.028}\次 {13.89 \压裂{千牛}{米}}\次{(6米)^ 2}= 14.001 千牛·米$$
• Negatives interior span:
• $$M_{x1,B} = -\lambda_{1X} \times M_x = -1.33 \次 19.512 kNm = – 25.951 kNm$$
• $$M_{y1,B} = -\lambda_{1和} \times M_y = -1.33 \次 14.001 kNm = – 18.621 千牛·米$$
• Negatives interior second span:
• $$M_{x2,B} = -\lambda_{2X} \times M_x = -1.33 \次 19.512 kNm = – 25.951 kNm$$
• $$M_{y2,B} = -\lambda_{2和} \times M_y = -1.33 \次 14.001 kNm = – 18.621 千牛·米$$

Rebar steel for X direction

$$\alpha$$ and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
M value 11.089 22.177 29.495 25.951 19.512 25.951
$$\rho_t$$ 0.00055614 0.00112 0.001496 0.001313 0.000984 0.001313
ku 0.015395 0.0310 0.0414 0.0364 0.0272 0.0364
$$\phi$$ 0.8 0.8 0.8 0.8 0.8 0.8
$$一个_{圣} {mm^2}$$ 334.8214 334.8214 335.08233 334.821 334.8214 334.8214

Rebar steel for Y direction

$$\alpha$$ and Moments Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
M value 8.751 17.501 23.276 18.621 14.001 18.621
$$\rho_t$$ 0.0004383 0.0008811 0.001176 0.0009381 0.000703 0.0009381
ku 0.0121 0.0244 0.03256 0.02597 0.0195 0.02597
$$\phi$$ 0.8 0.8 0.8 0.8 0.8 0.8
$$一个_{圣} {mm^2}$$ 334.821 334.821 334.821 334.821 334.8214 334.821

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### SkyCiv S3D Plate Design Module Results

After refining the model, is time to run a linear elastic analysis.

When designing slabs, we have to check if the vertical displacement are less than the maximum allowed by code. Australian Standars stablished a maximum serviciability vertical displacement of $$\压裂{大号}{250}= 分数{6000毫米}{250}=24.0 mm$$. The image above gaves to us the vertical displacement. The maximum value is -1.179mm being less than the maximum allowed of -24mm. 因此, the slab’s stiffeness is adequate. Images 27 和 28 consist of the bending moment in each main direction. Taking the moment distribution and values, the software, SkyCiv, can obtain then the total steel reinforcement area. Steel reinforcement areas:    ### Results comparison

The last step in this one-way slab design example is compare the steel rebar area obtained by S3D analysis and handcalculations.

Rebar steel for X direction

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$一个_{圣, HandCalcs} {mm^2}$$ 334.8214 334.8214 335.08233 334.821 334.8214 334.8214
$$一个_{圣, S3D} {mm^2}$$ 289.75 149.35 325.967 325.967 116.16 217.311
$$\三角洲_{dif}$$ (%) 13.461 55.39 2.720 2.644 65.307 35.0964

Rebar steel for Y direction

Moments and steel area Exterior Negative Left Exterior Positive Exterior Negative Right Interior Negative Left Interior Positive Interior Negative Right
$$一个_{圣, HandCalcs} {mm^2}$$ 334.821 334.821 334.821 334.821 334.821 334.821
$$一个_{圣, S3D} {mm^2}$$ 270.524 156.75 304.34 304.34 156.75 270.52
$$\三角洲_{dif}$$ (%) 19.203 53.184 9.104 9.104 53.184 19.204

The diference is some high for positive moments and the reason would be the presence of beams with high torsional stiffness that impact on the Plate Finite Element Analysis Results and the calculations for bending reinforcement steel.

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## 参考资料

1. 红豆杉 & Sanual Hug Chowdhury , “钢筋和预应力混凝土”, 2nd edition, Cambridge University Press.
2. Bazan Enrique & Meli Piralla, “Diseño Sísmico de Estructuras”, 1编辑, LIMUSA.
3. 澳大利亚标准, 混凝土结构, 如 3600:2018